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3 years ago

Hellllpppppppp helppppppppp

Chemistry

1 answer:

mixas84 [53]3 years ago

3 0

Answer :

Charles's Law : It is defined as the volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

Mathematically,

$V\propto T$

Boiling water bath Cool bath 1 Cool bath 2

Temperature (⁰C) 99 17 2

Temperature (K)(T) 273+99=372 273+17=290 273+2=275

Volume of water 0.0 27.0 34.0

in cool flask (mL)

Volume of water= 135.8 135.8 135.8

Air in flask (mL)

Volume of air 135.8 108.8 101.8

in cool flask (V)

$\frac{V}{T}$ $\frac{135.8}{372}=0.365$ $\frac{108.8}{290}=0.375$ $\frac{101.8}{275}=0.370$

The graph volume versus temperature for a gas is shown below.

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How much 10.0 M HNO must be added to 1.00 L of a buffer that is 0.0100 M acetic acid and 0.100 M sodium acetate to reduce the pH

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Explanation:

It is given that molarity of acetic acid = 0.0100 M

Therefore, moles of acetic acid = molarity of acetic acid × volume of buffer

Moles of acetic acid = 0.0100 M × 1.00 L

= 0.0100 mol

Similarly, moles of acetate = molarity of sodium acetat × volume of buffer

= 0.100 mol

When $HNO_{3}$ is added, it will convert acetate to acetic acid.

Hence, new moles acetic acid = (initial moles acetic acid) + (moles $HNO_{3}$ )

= 0.0100 mol + x

New moles of sodium acetate = (initial moles acetate) - (moles $HNO_{3}$ )

= 0.100 mol - x

According to Henderson - Hasselbalch equation,

pH = $pK_{a} + log\frac{[conjugate base]}{[weak acid]}$

pH = $pKa + log\frac{(new moles of sodium acetate)}{(new moles of acetic acid)}$

4.95 = 4.75 + $log\frac{(0.100 mol - x)}{(0.0100 mol + x)}$

$log\frac{(0.100 mol - x)}{(0.0100 mol + x)}$ = 4.95 - 4.75

= 0.20

$\frac{(0.100 mol - x)}{(0.0100 mol + x)}$ = antilog (0.20)

= 1.6

Hence, x = 0.032555 mol

Therefore, moles of $HNO_{3}$ = 0.032555 mol

volume of $HNO_{3}$ = $\frac{moles HNO_{3}}{molarity of HNO_{3}}$

= $\frac{0.032555 mol}{10.0 M}$

= 0.0032555 L

or, = 3.25 (as 1 L = 1000 mL)

Thus, we can conclude that volume of $HNO_{3}$ added is 3.26 mL.

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3 years ago